Moving rod viewed in different inertial frames

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1. Jun 26, 2016

Whitehole

1. The problem statement, all variables and given/known data
Two inertial frames S and S' are in standard configuration, the frame S' is moving along the x axis of S with velocity v. In S' a straight rod parallel to the x' axis moves in the y' direction with velocity u. Show that in S the rod is inclined to the x-axis at an angle $- tan^{-1}(\frac{γuv}{c^2})$

2. Relevant equations
$t = γ(t' + \frac{vx'}{c^2}) ~~~ x = γ(x' + vt') ~~~ Lorentz~~Transformation$
$γ = \frac{1}{[1 - (\frac{v}{c})^2]^½} ~~~ Lorentz~~ Factor$

3. The attempt at a solution
From the relativity of simultaneity, the front and back end of the rod will not be seen by the observer in S as having the same position as seen by the observer in S', the front end will lag behind the back end since light from the front will travel a little bit longer than the back to reach S.
I will treat the front and back end as separate case, let P1, P2 be the back and front end respectively. Place P1 at x' = 0 so that P2 is at x' = L

As seen by S, P1 is
$t = γ( t' + 0 ) = γt' ~~⇒~~ t' = \frac{t}{γ}$
$x = γ( 0 + vt' ) = γvt' = vt$
$y = y' = ut' = \frac{ut}{γ}$
$P1 = ( vt, \frac{ut}{γ} )$

As seen by S, P2 is
$t = γ( t' + \frac{vL}{c^2} )~~⇒~~ t' = \frac{t}{γ} - \frac{vL}{c^2}$
$x = γ( L + vt' ) = γL + γvt' = γL + γv( \frac{t}{γ} - \frac{vL}{c^2} ) = γL + vt - γL\frac{v^2}{c^2} = γL(1 - \frac{v^2}{c^2} ) + vt = \frac{L}{γ} + vt$
$y = y' = ut' = u( \frac{t}{γ} - \frac{vL}{c^2} ) = \frac{ut}{γ} - \frac{uvL}{c^2}$
$P2 = ( vt + \frac{L}{γ}, \frac{ut}{γ} - \frac{uvL}{c^2})$

$tan(θ) = \frac{Δy}{Δx} = \frac{ \frac{ut}{γ} - \frac{uvL}{c^2} - \frac{ut}{γ} }{ vt + \frac{L}{γ} - vt } = \frac{ - \frac{uvL}{c^2} }{ \frac{L}{γ} } = -γ\frac{uv}{c^2}$

$tan(-θ) = -tan(θ)~~~⇒~~~ tan(-θ) = γ\frac{uv}{c^2} ~~~⇒~~~ θ = -~ tan^{-1}(γ\frac{uv}{c^2})$

Is my solution correct?

2. Jun 27, 2016

Hamal_Arietis

yes, correct !